A) \[3x-y=0\]
B) \[x+3y=0\]
C) \[x-3y=0\]
D) \[3x+y=0\]
Correct Answer: C
Solution :
Given, centre of circle is\[(2,-1)\]and equation of tangent is \[3x+y=0\] \[\therefore \]Radius, \[r=\left| \frac{3(2)-1}{\sqrt{{{3}^{2}}+{{1}^{2}}}} \right|=\left| \frac{6-1}{\sqrt{10}} \right|=\frac{5}{\sqrt{10}}\] Now, equation of perpendicular line to the tangent line is \[x-3y=0\] Perpendicular distance of line\[x-3y=0\]from \[(2,-1)\]is \[p=\left| \frac{2-3(-1)}{\sqrt{{{1}^{2}}+{{3}^{2}}}} \right|=\left| \frac{2+3}{\sqrt{10}} \right|=\frac{5}{\sqrt{10}}=\]radius Hence, equation of second tangent is\[x-3y=0\].
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